Characterization of the non-negative definite functions $f(x,y)$

Question

The common definition of the non-negative definite functions is as follows:

Definition 1: A continuous complex-valued function $f(x)$ is called non-negative definite, if for any real numbers $x_1,\dots,x_m$ and complex numbers $\xi_1,\dots,\xi_m$, one has

$$ \sum_{k,j=1}^m f(x_k-x_j) \xi_k \bar{\xi}_j \ge 0\:. $$

See wikipedia for example.

For my use, I need the following more general definition:

Definition 2: A continuous complex-valued function $f(x,y)$ is called non-negative definite, if for any real numbers $x_1,\dots,x_m$ and complex numbers $\xi_1,\dots,\xi_m$, one has

$$ \sum_{k,j=1}^m f(x_k,x_j) \xi_k \bar{\xi}_j \ge 0\:. $$

For the first definition, there is a Characterization by Bochner's theorem, which is discussed in the post.

My question is, for the second definition, is there a characterization like Bochner's theorem? A necessary condition is that $f(x,y)$ is Hermitean in the sense that

$$ f(x,y) = \bar{f}(y,x)\:. $$

Clearly, we need more conditions to guarantee that $f(x,y)$ is non-negative definite. Does any one know a way to determine whether a given function $f(x,y)$ is non-negative definite or not?

Thank you very much for any hints!

Anand

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EDIT:

Thanks Mikael de la Salle for the useful link. After reading, I still have some problems.

As one might know, Definition 1 above can be extended to distributions. For example, a distribution $F$ is positive-definite, if for every $\psi\in C_c^\infty(R^d)$,

$$ \left(F,\psi *\widetilde{\psi}\right)\ge 0, $$

where $\widetilde{\psi}(x)=\psi(-x)$. So it is natural to think that Definition 2 can also be extended to, say, $C_c^\infty(R^d)\otimes C_c^\infty(R^d)$, or $C_c^\infty(R^{2d})$. The condition in the book Kazhdan's property (T) that the kernel function should be continuous seems too restrictive. Are there some more general statements?

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Second EDIT:

Here is a more explicit statement of the problem:

Theorem: Every translation-invariant positive-definite Hermitian bilinear functional $B(\phi,\psi)$ on $C_c^\infty(R^d)$ has the form

$$ B(\phi,\psi)= \int \hat{\phi}(\lambda) \overline{\hat{\psi}(\lambda)} d \mu(\lambda) $$

where $\mu$ is some positive tempered measure and $\hat{\psi}$, $\hat{\phi}$ are the Fourier transforms, respectively, of $\psi$ and $\phi$.

See Gel'fand Volumn 4 P.169 Theorem 6. . My problem is

   What happens if we remove the "translation--invariant" condition?

Thanks a lot!

Anand

Answer 1

Such a function is called function of positive type. For characterizations, see for example appendix C in the book Kazhdan's property (T) by Bekka, de la Harpe and Valette, available freely on Bachir Bekka's webpage.

Answer 2

On a Lie group, every positive-definite distribution is the distributional derivative of some continuous, positive-definite function. See Aarnes, Johan F. Distributions of positive type and representations of Lie groups. Math. Ann. 240 (1979), no. 2, 141–156.

Answer 3

In light of Anand's comments on translation invariant kernels of the form $f(x-y)$, I thought I'd mention the class of completely monotone functions $\left((-1)^k f^{(k)}(x) \ge 0\right)$---used in the famous Bernstein and Widder theorem

This theorem allows us to characterize completely monotone functions of the form $k(x,y) = f(x+y)$, by showing that equivalently they must be Laplace transforms of a positive measure.

Answer 4

It seems to me that if you drop the translation invariance condition on $B(\phi,\psi)$, then you simply end up with a symmetric quadratic form $B(\phi)=B(\phi,\phi)$. In that case the main simplification I can think of is the spectral representation: $$ B(\phi) = \int_{\operatorname{spec}(B)} ~ \lambda ~ \mathrm{d}\mu_\phi(\lambda) , $$ where $\operatorname{spec}(B)$ is the spectrum of $B$ and $\mathrm{d}\mu_\phi$ is the spectral measure associated to $\phi$.

The spectral theory of quadratic forms is discussed in section VIII.6 of Reed & Simon's Methods of Modern Mathematical Physics (vol. 1). The main theorem is (VIII.15): If $B$ is a semibounded quadratic form, then it is the quadratic form $B(\phi)=(\phi,A\phi)$ of a unique self-adjoint operator $A$. The inner product in your case is the obvious one on $L^2(\mathbb{R})$ and the spectral representation of $B$ is obtained directly from the spectral representation of $A$. Also, since your $B$ is positive semidefinite, the spectrum will also be bounded $\operatorname{spec}(B)\ge 0$.

When $B$ is translation invariant, the unitary transformation that takes $A$ to its diagonal form is automatically the Fourier transform and $\mathrm{d}\mu_\phi(\lambda) = |\hat{\phi}(\lambda)|^2\mathrm{d}\lambda$ (well, up to a measurable reparametrization of $\lambda$ really), where $\hat{\phi}(\lambda)$ is the Fourier transform of $\phi(x)$. In the generic case, short of explicitly diagonalizing $B$ or $A$, I don't see a simple way of obtaining $\mathrm{d}\mu_\phi$.